Let there be two functions $f,g:(a,b) \to\mathbb R$ that are differentiable in $(a,b)$ with either
$$\text{Case 1:}\qquad\lim_{x\to b} f(x) = \lim_{x\to b} g(x) = 0$$ or $$\,\,\,\,\,\,\,\text{Case 2:}\qquad\lim_{x\to b} f(x) = \lim_{x\to b} g(x) = \pm \infty$$

I want to show that $$\lim_{x\to b} \frac{f(x)}{g(x)} = \lim_{x\to b} \frac{f'(x)}{g'(x)}$$ if the right hand side of the equation exists.

My approach:
Since $f,g$ are differentiable, we have $f'(x) \approx \frac{f(x+d)-f(x)}d$, where $d$ is an infinitesimal.
Per algebraic transformations we can therefore deduce: $$f'(x) \approx \frac{f(x+d)-f(x)}d \\\Leftrightarrow\\ f'(x) +d_f =\frac{f(x+d)-f(x)}d \\\Leftrightarrow\\ f(x+d) = f(x) + df'(x) +dd_f $$ (Which is a nice identity by itself)

From now on, let every variable of the form $d_x$ be an infinitesimal. We can deduce: $$\lim_{x\to b} \frac{f(x)}{g(x)} \approx\frac{f(b+d+d_2)}{g(b+d+d_2)} =\frac{f(b+d) + d_2f'(b+d) +d_2d_f}{g(b+d) + d_2g'(b+d) +d_2d_g} $$ (I've used $d+d_2$ rather than just $d$ because otherwise $f'(b)$ would be undefined)

Case 1: If the limit tends to $0$, we can simplify: $$\frac{f(b+d) + d_2f'(b+d) +d_2d_f}{g(b+d) + d_2g'(b+d) +d_2d_g} = \frac{d_3 + d_2f'(b+d) +d_2d_f}{d_4 + d_2g'(b+d) +d_2d_g} $$

And here I'm stuck. $d_3$ and $d_4$ make it impossible to cancel out $d_2$.

Is there a way to make my reasoning whole?

  • $\begingroup$ @DarkKnight As a derivative is a limit process, the intuitive nonstandard-definition of the derivative is as I have given it (i.e. with $\approx$). However, I'm struggling to find a counterexample that clearly shows that $=$ can not be the case, so maybe this is a possible definition as well $\endgroup$ – Sudix Jun 15 '18 at 3:44
  • $\begingroup$ @DarkKnight I found one after all: If $f'(x) = \dfrac{f(x+d)-f(x)}{d}$, then this means the equation holds for all infinitesimal $d$. Therefore, $$\dfrac{f(x+d)-f(x)}{d} = \dfrac{f(x+d_2)-f(x)}{d_2}$$. So, if we set $f(x) = x^2: $$$\dfrac{2xd+d^2}{d} = \dfrac{2xd_2+{d_2}^2}{d_2} \Leftrightarrow \frac{x^2}d - \frac{x^2}{d_2} = d-d_2$$ You arrive at a contradiction if you substitute x for any value and solve for $d$. You then see, that the equation only holds for '''some''' choices of $d,d_2$, not for '''all''. $\endgroup$ – Sudix Jun 15 '18 at 3:55
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    $\begingroup$ Why is $\lim_{x\to b} \frac{f(x)}{g(x)} \approx\frac{f(b+d+d_2)}{g(b+d+d_2)}$? $\endgroup$ – Mikhail Katz Jun 15 '18 at 7:12
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    $\begingroup$ The rule you stated is correct under the assumption that the denominator is appreciable (not infinitesimal). But here the denominator is infinitesimal so the rule is not applicable. $\endgroup$ – Mikhail Katz Jun 17 '18 at 7:58
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    $\begingroup$ The case of L'Hopital is proved in Keisler's book on page 243 is when the limit is $f'(c)/g'(c)$ but the general case is probably treated in the accompanying "foundations" book. I didn't check though. $\endgroup$ – Mikhail Katz Jun 18 '18 at 13:48

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